February 18 1 Reaction Mechanism 14.4 Reaction Mechanism Steps of a Reaction Fred Omega Garces Chemistry 201 Miramar College
February 181 Reaction Mechanism
14.4 Reaction Mechanism
Steps of a Reaction
Fred Omega GarcesChemistry 201Miramar College
February 182 Reaction Mechanism
The Ozone LayerOzone is most important in the stratosphere, at this level in the atmosphere, ozone absorbs UV radiation
10 Km
50 Km
100 Km
Ozone Layer
Stratosphere
Mesosphere
TroposphereMt. Everest
TroposphereStratosphereMesosphere
February 183 Reaction Mechanism
Mechanism of Ozone; Chapman Cycle
Chapman cycle shows that O3 exist at steady state. It is constant in the stratosphere.
O
O2
O3
O
O2hν
hν
O2
O2
O3
2 O
O
320 nm or less
242 nm or less
Slow ozone removal step
O3 lives for ~200 - 300 s before it dissociates.Ozone removal step:O3 + O g 2O2
February 184 Reaction Mechanism
Reaction Coordinate for ozone destruction
A Key reaction in the upper atmosphere is
O3 (g) + O (g)® 2O2 (g)
The Ea (fwd) is 19 kJ, and the DHrxn as written is -392 kJ.
A reaction energy diagram for this reaction with the calculate Ea(rev) is shown.
O3 + O
2O2
O kJ
19 kJ
-392 kJ
Reaction Progress
Eact (rev)= 411 kJ
February 185 Reaction Mechanism
1. Water Vapors: H2O OH• + H•H• + O3 OH• + O2OH• + O H• + O2
Net: O + O3 2O2
2. N2 , Dinitrogen : N2 + O2 2NONO + O3 NO2 + O2NO2 + O NO + O2
Net: O + O3 2O2
3. CFCs CCl2F2 CClF2• + Cl•Chlorofluorocarbons Cl• + O3 ClO• + O2
ClO• + O Cl• + O2Net: O + O3 2O2
Path to Destruction: Ozone
\ 10,000 O3 will breakdown to O2 for every Cl•
February 186 Reaction Mechanism
Influence by CFC: OzoneComparison of activation energies in the uncatalyzed decompositions of ozone. The destruction of ozone can be catalyzed by Cl atoms which leads to an alternative pathway with lower activation energy, and therefore a faster reaction.
Progress of reaction
Ener
gy (k
J)
February 187 Reaction Mechanism
Reaction MechanismThe mechanism of a reaction is the sequence of steps (at the molecular level) that shows how reactant chemicals combine to form the final products.
Elementary StepsSequence of steps which describes an actual molecular event.
StoichiometryThe overall stoichiometric reaction is the sum of the elementary steps.
Scientist want to learn about mechanisms because an understanding of the
mechanism (how bonds break and form) may lead to conditions to improve reaction
product yield, (or prevent side products formation. i.e, depletion of ozone.)
February 188 Reaction Mechanism
Ozone: Revisited
Chapman�s CycleO3 ® O2 + OO + O3 ® 2O22O3 ® 3O2
Consider an elementary step
iA + jB ® Product (slow step)
rate = k [A]i •[B]j
The rate of the reaction is directly proportional to concentrations of the colliding species.
Elementary Steps give rise to Rate LawSince elementary steps describes a
molecular collision, the rate law for an
elementary step (unlike the overall reaction)
can be written from the Stoichiometry.
February 189 Reaction Mechanism
Elementary Step: Rate LawConsider the following proposed mechanism for the conversion of NO2 to N2O5. What is the rate law.Step1 NO2 + O3 ® NO3 + O2 (slow)Step2 NO3 + NO2 ® N2O5 (fast)
rate = k1[NO2]1 [O3]1
In a series of steps, the slowest step determines the overall rate.
In the mechanism for a chemical reaction, the slowest step is the rate-determining step.
February 1810 Reaction Mechanism
Elementary Steps: Order of reactionElem. Step Rate Law Order Molecularity
1 A ® Product Rate = k[A] 1st order unimolecular
2 2A ® Product Rate = k[A]2 2nd order bimolecularA + B ® Prod. Rate = k[A][B]
3 3A ® Product Rate = k[A]3 3rd order Termolecular2A + B ® Product Rate = k[A]2[B]A + B + C ® Prod Rate = k[A][B][C]
* Termolecular mechanism (3-elementary step) is very rare.Scientist who propose such a mechanism must make careful measurements.
February 1811 Reaction Mechanism
Multiple Elementary Steps Most reactions involve more than one elementary step.
Rate-Limiting - When one step is much slower than any other, the overall rate is determined by the slowest �Rate-determining� step.
Reaction is only as fast as the slowest elementary step
Analogy: Leaving class after an exam.
On a single lane highway, speed of traffic is only as fast as creepy crawler 12-cars ahead.
February 1812 Reaction Mechanism
Rate Determining Step from Rate Law
Consider: NO2 + CO g NO + CO2
Mechanism: (1) NO2 (g) + NO2 (g) ® NO3 g) + NO (g)
(2) NO3 (g) + CO (g) ® NO2 + CO2 (g)
Net: NO2 (g) + CO (g) ® CO2 (g ) + NO (g)
Rate = k[NO2]2
Which is Rate limiting step (1) or (2) ?RDS is the step that determines the rate law.
When scientists propose a mechanism, they can only say that it is consistent with the experimental data.
There may be other mechanisms that are consistent with experimental data as well.
If experiments are done in the future to disprove the mechanism, then the proposed mechanism must be revised.
February 1813 Reaction Mechanism
RDS and Rate Law: ExampleConsider the reaction : NO (g) + O3 ® NO2 + O2Two mechanisms (elementary steps) are proposed:
Mechanism 1 NO + O3 ® NO2 + O2Proposed rate law: Rate = k [NO] [O3]
Mechanism 2 O3 ® O2 + O (slow)NO + O ® NO2 (fast)
Proposed rate law: Rate = K [O3]
What are the Rate Laws ?When a potential mechanism is proposed, 2 factors must be considered -
•Rate limiting step must be consistent with observed rate law.•Sum of all the steps must yield the observed stoichiometry.
February 1814 Reaction Mechanism
Complicated Reaction MechanismReaction mechanism in which slow step (rate determining step) involves an intermediate.
Consider: A ® BMechanism: A D int (fast)
int ® B (slow)NET: A ® B
RATE = k[int]-but the rate law cannot be written in terms of an intermediate(catalyst and reactant okay, but not intermediate).
-It must be expressed in terms of stable speciesHow is the Rate Law modified?
February 1815 Reaction Mechanism
Modification of Rate Law
RATE = k [int]Written in terms of reactants-
The rate law is now expressed in terms of the reactant.
€
Rate = k [int]
keq =[int][A]
→ [int] = keq[A]
Rate = k∗keq[A] = K' [A]
February 1816 Reaction Mechanism
Rate Laws from Mult. Steps Mechanism.Consider the reaction below, what is the rate law based on the two
proposed mechanism:2 NO2 (g) + O3 ® N2O5 + O2
Mechanism (1) Mechanism (2)NO2 + NO2 D N2O4 (fast) NO2 + O3 D NO3 + O2 (slow)N2O4 + O3 ® N2O5 + O2 (slow) NO3 + NO2 ® N2O5 (fast)
Rate = k [N2O
4] [O
3]
keq=
[N2O
4]
[NO2] [NO
2]
[N2O
4] = k
eq[NO
2]2
Rate = k ∗keq
[NO2]2[O
3]
Rate = k[NO2]2[O
3]
Termolecular
€
Rate = k[NO2] [O3]bimolecular
Rate is based onslowest elem. step
February 1817 Reaction Mechanism
Rate Laws from Mult. Steps Mechanism.The decomposition of hydrogen peroxide is catalyzed by iodide ion. The
catalyzed reaction is thought to proceed b a two-step mechanism:
H2O2 (aq) + I-(aq) ® H2O(l) + IO-
(aq) (slow)H2O2 (aq) + IO-
(aq) ® H2O(l) + I-(aq) + O2 (g) (fast)
a) Rate Law:Rate Law = k [H2O2] [I- ]
b) Overall reaction:2 H2O2 (aq) ® H2O(l) + O2 (g)
c) Intermediate: IO-(aq)
Catalyst: I-(aq)
February 1818 Reaction Mechanism
Enzyme Catalysis Reaction
Consider oxidation of ethanol to aldehyde:CH3CH2OH (l) ® ADH ® CH3CHO + R-H2
ADH - Alcohol dehydrogenaseMechanism
E + S D ESES ® E + P (slow)
E + S ® E + P
Rate = k [ES]
Keq = [ES] ® Keq [E] [S] = [ES][E] [S]
Rate = K Keq [E] [S] = K� [E] [S]
February 1820 Reaction Mechanism
In Class Exercise
Phosgene (Cl2CO) a poison gas used in WW1, is formed in the following reaction.Mechanism: i) H+ + Cl2 D Cl + HCl fast, reversible
ii) HCl + CO D ClCO + H+ fast, reversible
iii) ClCO + Cl® Cl2CO slow
i) What is the overall reaction?
ii) What is the intermediates and the catalyst for the reaction?
iii) What is the rate law from this mechanism?
iv)What is the order of the reaction with respect to each reactant?
v) What is the overall molecularity?
vi) How would doubling the concentration of CO affect the reaction?
February 1821 Reaction Mechanism
Summary
The dynamics of the series of steps of a chemical change is what
kinetics tries to explain. Variation in reaction rate are observed
through concentration and temperature changes, which operate on
the molecular level through the energy of particle collision. Kinetics
allows us to speculate about the molecular pathway of a reaction.
Modern industry and biochemistry depend on its principles.
However, speed and yield are very different aspects of a reaction.
Speed is in the kinetic domain, yield is in the equilibrium domain.